[PD] number to fractions external?
Jonathan Wilkes
jancsika at yahoo.com
Fri Dec 16 20:46:24 CET 2011
>________________________________
> From: i go bananas <hard.off at gmail.com>
>To: Ludwig Maes <ludwig.maes at gmail.com>
>Cc: pd-list at iem.at
>Sent: Friday, December 16, 2011 1:16 PM
>Subject: Re: [PD] number to fractions external?
>
>
>if you had read the thread, you would have seen that claude posted a link to that technique.
>
>now go and make a PD patch that does it, mr smart guy.
Wow, how much cpu does that take in Python? I tried this approach in the form
of an abstraction, with a nested until, and worst case it can take as much as a quarter of
a second to compute with the constants provided below.
(Pentium Dual-core 2.6gHz in WinXP with 0.43 nightly build)
-Jonathan
>
>
>
>
>On Sat, Dec 17, 2011 at 3:00 AM, Ludwig Maes <ludwig.maes at gmail.com> wrote:
>
>If you guys 'd done your math, you'd know there is an ancient algorithm for approximating numbers by fractions and its called continued fractions.
>>
>>
>>
>>On 16 December 2011 18:38, Lorenzo Sutton <lorenzofsutton at gmail.com> wrote:
>>
>>On 16/12/11 16:05, Alexandre Torres Porres wrote:
>>>
>>>looks like a job for an external
>>>>
Not really answering the OP question but something could be done in Python:
>>>
>>>def find_frac(num):
>>> f = float(num)
>>> last_error = 1000
>>> best = (0,0)
>>> for i in xrange(1,1001):
>>> for j in xrange(1,i+1):
>>> divide = (float(i) / float (j))
>>> if divide == f:
>>> return ((i,j),0)
>>> err = abs(divide - f)
>>> if err < last_error:
>>> best = (i,j)
>>> last_error = err
>>> return (best,last_error)
>>>
>>>This would try to find the exact fraction or the one with the smallest error (trying up to 1000/1000). It would return (numerator, denominator, error). Guess it would work well at least up to 100 but only for positive numbers... and... not for numbers < 1.. and surely it's not optimised etc. etc. :)
>>>
>>>>>> find_frac(2)
>>>((2, 1), 0)
>>>>>> find_frac(1.5)
>>>((3, 2), 0)
>>>>>> find_frac(1.333333333333333333333333333)
>>>((4, 3), 0)
>>>>>> find_frac(2.4)
>>>((12, 5), 0)
>>>>>> find_frac(2.8)
>>>((14, 5), 0)
>>>>>> find_frac(2.987654321)
>>>((242, 81), 1.234568003383174e-11)
>>>>>> find_frac(50.32)
>>>((956, 19), 0.004210526315787888)
>>>>>> find_frac(50.322)
>>>((956, 19), 0.006210526315790332)
>>>>>> find_frac(50.4)
>>>((252, 5), 0)
>>>>>> find_frac(10.33)
>>>((971, 94), 0.00021276595744623705)
>>>>>> find_frac(10.33333333333333333333333333)
>>>((31, 3), 0)
>>>
>>>Lorenzo.
>>>
>>>
>>>>
>>>>
>>>>2011/12/16 i go bananas <hard.off at gmail.com <mailto:hard.off at gmail.com>>
>>>>
>>>>
>>>> actually, i'm not going to do anything more on this.
>>>>
>>>> i had a look at the articles claude posted, and they went a bit
>>>> far over my head.
>>>>
>>>> my patch will still work for basic things like 1/4 and 7/8, but i
>>>> wouldn't depend on it working for a serious application. As you
>>>> first suggested, it's not so simple, and if you read claude's
>>>> articles, you will see that it isn't.
>>>>
>>>> it's not brain science though, so maybe someone with a bit more
>>>> number understanding can tackle it.
>>>>
>>>>
>>>>
>>>> On Sat, Dec 17, 2011 at 12:51 AM, Alexandre Torres Porres
>>>>
>>>> <porres at gmail.com <mailto:porres at gmail.com>> wrote:
>>>>
>>>> > i had a go at it
>>>>
>>>> thanks, I kinda had to go too, but no time... :(
>>>>
>>>> > yeah, my patch only works for rational numbers.
>>>>
>>>> you know what, I think I asked this before on this list,
>>>>
>>>> deja'vu
>>>>
>>>> > will have a look at the article / method you posted, claude.
>>>>
>>>> are you going at it too? :)
>>>>
>>>> by the way, I meant something like 1.75 becomes 7/4 and not
>>>> 3/4, but that is easy to adapt on your patch
>>>>
>>>> thanks
>>>>
>>>> cheers
>>>>
>>>>
>>>>
>>>> 2011/12/16 i go bananas <hard.off at gmail.com
>>>>
<mailto:hard.off at gmail.com>>
>>>>
>>>>
>>>> by the way, here is the method i used:
>>>>
>>>> first, convert the decimal part to a fraction in the form
>>>> of n/100000
>>>> next, find the highest common factor of n and 100000
>>>> (using the 'division method' like this:
>>>> http://easycalculation.com/what-is-hcf.php )
>>>>
>>>> then just divide n and 100000 by that factor.
>>>>
>>>> actually, that means it's accurate to 6 decimal places, i
>>>> guess. well...whatever :D
>>>>
>>>>
>>>>
>>>>
>>>>
>>>>
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>>>
>>>
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